Question

A sample of a gas occupies a volume of ${\text{512}}$mL at ${\text{2}}{{\text{0}}^{\text{o}}}{\text{C}}$and $74$cm of Hg as pressure. What volume would this gas occupy at STP?

Answer 1

Hint: To answer this question we should know the combined gas law. According to this law, if an ideal gas is present at a condition of temperature, pressure, and volume, and any one or two parameters of temperature, pressure, or volume get changed then we can determine the third parameter by putting the initial conditions of temperature pressure and volume equal to the final condition of temperature, pressure, and volume.

Formula used: $\dfrac{{{{\text{p}}_{\text{1}}}{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}}\, = \,\dfrac{{{{\text{p}}_2}{{\text{V}}_2}}}{{{{\text{T}}_2}}}$

Complete step-by-step Solution:
The relation between temperature pressure and volume of an ideal gas according to combined gas law is as follows:
$\dfrac{{{{\text{p}}_{\text{1}}}{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}}\, = \,\dfrac{{{{\text{p}}_2}{{\text{V}}_2}}}{{{{\text{T}}_2}}}$
${{\text{p}}_{\text{1}}}$is the initial pressure
${{\text{V}}_{\text{1}}}$is the initial volume
${{\text{T}}_{\text{1}}}$is the initial temperature
${{\text{p}}_2}$is the final pressure
${{\text{V}}_2}$is the final volume
${{\text{T}}_2}$ is the final temperature
Initial pressure is$74$cm of Hg. We will convert the pressure from cm of Hg to atm as follows:
$76$ cm of Hg = $1$ atm
$74$cm of Hg = $1.02$ atm
Initial temperature is${\text{2}}{{\text{0}}^{\text{o}}}{\text{C}}$. We will convert the temperature from degree Celsius to kelvin as follows:
${\text{K}}\,{\text{ = }}{\,^{\text{o}}}{\text{C}}\,{\text{ + }}\,{\text{273}}\,$
${\text{K}}\,{\text{ = }}\,{20^{\text{o}}}{\text{C}}\,{\text{ + }}\,{\text{273}}\,$
$\,{\text{293}}\,{\text{K}}\,$
So, Initial volume is ${\text{512}}$mL, pressure is $1.02$ atm and temperature is $\,{\text{293}}\,{\text{K}}\,$.
The $273$K temperature and $1$ atm pressure is known as standard temperature and pressure so, the final temperature is $273$and pressure is$1$ atm.

By using combined law we will calculate the final volume at STP as follows:
On substituting for$1.02$ atm ${{\text{p}}_{\text{1}}}$, ${\text{512}}$mL for${{\text{V}}_{\text{1}}}$, $\,{\text{293}}\,{\text{K}}\,$for ${{\text{T}}_{\text{1}}}$, $273$for${{\text{T}}_2}$, $1$atm for ${{\text{p}}_2}$,
$\dfrac{{{\text{1}}{\text{.02}}\,{\text{atm}}\, \times 512\,{\text{ml}}}}{{\,{\text{293}}\,{\text{K}}\,}}\, = \,\dfrac{{{\text{1}}\,{\text{atm}}\, \times {{\text{V}}_2}}}{{\,{\text{273}}\,{\text{K}}}}$
\[{{\text{V}}_2}\, = \,\,\dfrac{{{\text{1}}{\text{.02}}\,{\text{atm}}\, \times 512\,{\text{ml}} \times {\text{273}}\,{\text{K}}}}{{\,{\text{1}}\,{\text{atm}}\, \times {\text{293}}\,{\text{K}}\,}}\,\]
\[{{\text{V}}_2}\, = \,\,\dfrac{{142571.52\,{\text{ml}}}}{{{\text{293}}\,\,}}\,\]
\[{{\text{V}}_2}\, = \,\,487\,{\text{ml}}\,\]
So, the volume occupied by the given gas at STP is\[487\,\]ml.

Therefore, \[487\,\]ml is the correct answer.

Note: The ideal gas equation is, ${\text{pV}}\,{\text{ = }}\,{\text{nRT}}$. Ideal gas law is a combination of three laws. Boyle law, according to that at constant temperature, pressure is inversely proportional to the volume. Charles’s law, at constant pressure, volume is directly proportional to the temperature. Avogadro’s law, according to that at constant temperature and pressure, volume is directly proportional to the number of moles. For the same gas with same number of moles, the R and n become constant so, we can write the ideal as equation as $\dfrac{{{\text{pV}}}}{{\text{T}}}\,{\text{ = }}\,{\text{nR}}$. When we compare this equation at two different conditions then we get $\dfrac{{{{\text{p}}_{\text{1}}}{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}}\, = \,\dfrac{{{{\text{p}}_2}{{\text{V}}_2}}}{{{{\text{T}}_2}}}$.